Solve x^2+left(40-xright)^2=818

$${ x }^{ 2 } + { \left(40-x \right) }^{ 2 } =818$$

Solve for x

$$x=17$$

$$x=23$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(40-x\right)^{2}.

$$x^{2}+1600-80x+x^{2}=818$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}+1600-80x=818$$

Subtract 818 from both sides.

$$2x^{2}+1600-80x-818=0$$

Subtract 818 from 1600 to get 782.

$$2x^{2}+782-80x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$2x^{2}-80x+782=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -80 for b, and 782 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-80\right)±\sqrt{\left(-80\right)^{2}-4\times 2\times 782}}{2\times 2}$$

Square -80.

$$x=\frac{-\left(-80\right)±\sqrt{6400-4\times 2\times 782}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-80\right)±\sqrt{6400-8\times 782}}{2\times 2}$$

Multiply -8 times 782.

$$x=\frac{-\left(-80\right)±\sqrt{6400-6256}}{2\times 2}$$

Add 6400 to -6256.

$$x=\frac{-\left(-80\right)±\sqrt{144}}{2\times 2}$$

Take the square root of 144.

$$x=\frac{-\left(-80\right)±12}{2\times 2}$$

The opposite of -80 is 80.

$$x=\frac{80±12}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{80±12}{4}$$

Now solve the equation x=\frac{80±12}{4} when ± is plus. Add 80 to 12.

$$x=\frac{92}{4}$$

Divide 92 by 4.

$$x=23$$

Now solve the equation x=\frac{80±12}{4} when ± is minus. Subtract 12 from 80.

$$x=\frac{68}{4}$$

Divide 68 by 4.

$$x=17$$

The equation is now solved.

$$x=23 x=17$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(40-x\right)^{2}.

$$x^{2}+1600-80x+x^{2}=818$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}+1600-80x=818$$

Subtract 1600 from both sides.

$$2x^{2}-80x=818-1600$$

Subtract 1600 from 818 to get -782.

$$2x^{2}-80x=-782$$

Divide both sides by 2.

$$\frac{2x^{2}-80x}{2}=-\frac{782}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{80}{2}\right)x=-\frac{782}{2}$$

Divide -80 by 2.

$$x^{2}-40x=-\frac{782}{2}$$

Divide -782 by 2.

$$x^{2}-40x=-391$$

Divide -40, the coefficient of the x term, by 2 to get -20. Then add the square of -20 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-40x+\left(-20\right)^{2}=-391+\left(-20\right)^{2}$$

Square -20.

$$x^{2}-40x+400=-391+400$$

Add -391 to 400.

$$x^{2}-40x+400=9$$

Factor x^{2}-40x+400. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-20\right)^{2}=9$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-20\right)^{2}}=\sqrt{9}$$

Simplify.

$$x-20=3 x-20=-3$$

Add 20 to both sides of the equation.

$$x=23 x=17$$